Question

Last question of the day :-

Evaluate :-
$\sf \dfrac{ {sin}^{2} 63 + {sin}^{2}27 }{cos {}^{2} 17 + {cos}^{2} 73}$
​

Answer 1

$\large\bf{\underline{\underline{Answer}}}$

$\boxed{\red{\sf\frac{sin}^{2}63° + sin² 27°}{cos² 17° + cos² 73°}}}$

$\sf{= sin² 63° + sin² (90° - 63°)}$

$\sf{cos² 17° + cos² (90° - 17°)}$

$\sf{sin² 63° + cos² 63°}$

$\sf{cos² 17° + sin² 17°}$

$\underline{\boxed{\sf\purple{= 1}}}$

$\sf{∵ sin(90° - θ) = cosθ;}$

$\sf{cos(90° - θ) = sinθ;}$

$\red{\bf {sin²θ + cos² θ = 1}}$

Thankyou :)

Answer 2

Given :-

$\longmapsto \sf\bold{\pink{\dfrac{{sin}^{2}63 + {sin}^{2}27}{{cos}^{2}17 + {cos}^{2}73}}}$

Solution :-

$\leadsto \sf \dfrac{{sin}^{2}63 + {sin}^{2}27}{{cos}^{2}17 + {cos}^{2}73}$

$\implies \sf \dfrac{{sin}^{2}63 + {sin}^{2}(90 - 63)}{{cos}^{2}17 + {cos}^{2}(90 - 17)}$

As we know that,

$\mapsto \sf sin(90 - \theta) =\: cos\theta$

$\mapsto \sf cos(90 - \theta) =\: sin\theta$

$\implies \sf \dfrac{{sin}^{2}63 + {cos}^{2}63}{{cos}^{2}17 + {sin}^{2}17}$

Again, as we know that,

$\mapsto \sf {sin}^{2}A + {cos}^{2}A =\: 1$

$\implies \sf \dfrac{\cancel{1}}{\cancel{1}}$

$\implies \sf\boxed{\bold{\red{1}}}$

$\sf\boxed{\bold{\green{\therefore\: The\: answer\: is\: 1.}}}$

$\rule{150}{2}$

Some Important Formula :

$\mapsto \sf\bold{\purple{cos(90 - \theta) =\: sin\theta}}$

$\mapsto \sf\bold{\purple{ sec(90 - \theta) =\: cosec\theta}}$

$\mapsto \sf\bold{\purple{tan(90 - \theta) =\: cot\theta}}$

$\mapsto \sf\bold{\purple{cosec(90 - \theta) =\: sec\theta}}$

$\mapsto \sf\bold{\purple{sin(90 - \theta) =\: cos\theta}}$

$\mapsto \sf\bold{\purple{cot(90 - \theta) =\: tan\theta}}$